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Question: What is the angle between \(C \times d\) & \(\hat C - \hat d\)?...

What is the angle between C×dC \times d & C^−d^\hat C - \hat d?

Explanation

Solution

Here we have to apply the concept of vectors. The cross product of two vectors is called a vector which has a direction perpendicular to the plane having the two vectors. If two vectors are having the same magnitude and same direction then they are considered to be equal.

Complete step-by-step solution:
The angle between two vectors is the smallest angle at which any two vectors are rotated about the other vector here, with both of the vectors having the same direction.”
From the right-hand rule of cross-product if we put our palm on the first vector and curl the fingers towards the other one then the resultant vector will be towards the thumb pointing.
The magnitude of the cross product is equal to the product of the magnitudes of the two vectors multiplied times the size of the angle between them to Compute its magnitude.
Now,
The dot product of C×dC \times d & C^−d^\hat C - \hat d
(C×d).(C^−d^)(C \times d).(\hat C - \hat d)
Expand the above equation,
(C×d).(C^−d^)=(C^×d^).C^−(C^−d^)d^(C \times d).(\hat C - \hat d) = (\hat C \times \hat d).\hat C - (\hat C - \hat d)\hat d
⇒(C×d).(C^−d^)=0−0\Rightarrow (C \times d).(\hat C - \hat d) = 0 - 0
⇒(C×d).(C^−d^)=0\Rightarrow (C \times d).(\hat C - \hat d) = 0
Since the dot product is zero. That means they are perpendicular to each other,
∴θ=900\therefore \theta = {90^0}
Finally, The angle between C×dC \times d & C^−d^\hat C - \hat d is θ=900\theta = {90^0}
If two vectors are having the same magnitude and same direction then they are considered to be equal.
When the two vectors having equal magnitude but opposite in direction they are known as opposite vectors

Note: “Perpendicular" refers to the angle between the two vectors is 9090 degrees.
To identify the two vectors are perpendicular or not, we have to take the cross product of them;
When the cross product is equal to zero, then the vectors are perpendicular.